[seqfan] Re: slowest-growing "accelerating" sequence

Jon Wild wild at music.mcgill.ca
Fri Sep 13 17:48:42 CEST 2013


I don't seem to have received Don's comment, but I see it quoted below in 
Allan Wechsler's post. Don is correct that I meant for a(n) to be strictly 
increasing, too. At first I wrote the definition to make explicit that it 
was the slowest-growing sequence that started "1, 2...", then realised 
this was redundant if a(0) is also constrained to be strictly increasing.

Thank you for the extra terms Don! I had wondered how the overall growth, 
with its accelerating acceleration, would compare to exponential and other 
functions. By the way I also have the corresponding sequence using 
exponentiation instead of multiplication/division between successive 
terms--I'll submit that one too. Of course it grows even more quickly.

Allan, you are definitely correct that one can make sequences with much 
lower values (and hence slower growth) further in the sequence, by 
changing the starting values of the sequence. I couldn't see how to pick 
out any one of these as significant, though.

Jon

On Fri, 13 Sep 2013, Allan Wechsler wrote:

> I am intrigued by Don Reble's comment about the similar sequence beginning
> (1, 1, ,,,) growing faster than Wild's sequence starting (1, 2, ...). This
> suggests that the "greedy" algorithm of selecting the smallest next number
> that satisfies the constraints does not minimize subsequent entries. Is it
> possible that starting (1, 3, ...) makes subsequent entries even smaller?
> Or perhaps later non-optimal choices can reduce even later entries?
>
>
> On Fri, Sep 13, 2013 at 8:49 AM, Don Reble <djr at nk.ca> wrote:
>
>> a(n) = 1,2,5,16,68,404,3587,51747,**1343181,70863530...
>>>
>>> ... a(n) is the slowest growing integer sequence beginning with 1
>>>
>>> whose sequences of mth-order quotients are all strictly increasing,
>>> for all values of m.
>>>
>>> jon wild
>>>
>>
>>    Perhaps Jon means _all_ values of m, including 0.
>>    (So that it can't begin 1,1.)  Or perhaps not,
>>        1 1 2  9 103 3405 371146 153037899 275315985020 2505668630734788
>>        1 2 5 16  68  404   3587     51747      1343181         70863530
>>    because the 1,1 sequence grows faster anyway. Jon?
>>
>>    BTW, corresponding difference sequences are A000079 and A000225.
>>
>>  ...take the product of the previous element a(n) and all of the most
>>>
>>> recently calculated mth ratios, add 1, and take the floor.
>>>
>>> --Graeme McRae
>>>
>>
>>    Yes. The growth rate is hyperexponential, so my computer runs out of
>>    memory pretty quickly. (Those quotients are killers, not the sequence
>>    itself.) Have s'more terms.
>>
>> 1
>> 2
>> 5
>> 16
>> 68
>> 404
>> 3587
>> 51747
>> 1343181
>> 70863530
>> 8771738549
>> 3010577465956
>> 3472890614331263
>> 16778636239137208665
>> 435866021017987040200927
>> 80724349462285706161330870948
>> 146347653618405076026597153184**017178
>> 370281025373267805703902488419**0121454618096
>> 194146390420750116019045958924**0443604643356373706327
>> 327264872668416385059974502188**994396154497307865899952930516**57
>> 288446581117129009565353747088**521585632897325578890700993393**
>> 37027912646286
>> 227469456701027991010652297470**211018056216119398577688089952**
>> 1251905648932937035326556839
>> 290075547022486333813693535531**230950983289112758751771686880**
>> 396844514243266765608885248344**78778858475108
>> 114654236288391623803120048812**338378923201909025140523249822**
>> 058948949003096097601289509973**594977341379753585511691703312**066
>> 286771527322364315637265554893**187362324790934083293394580448**
>> 968848634100735335827561955127**136999494818157671130578737141**
>> 568527819282103471406137
>> 991640462838273375627738696091**672948855453024266252499046892**
>> 313948814248377868053876458901**350496387620054463160116560544**
>> 010813779680789761429009802800**978591899964162597
>>
>> --
>> Don Reble  djr at nk.ca
>>
>>
>>
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